Mark the correct alternative in each of the following:
Let g(x) = 1 + x – [x] and where [x] denotes the greatest integer less than or equal to x. Then for all x, f (g(x)) is equal to
Given that g(x) = 1 + x – [x] and
where [x] denotes the greatest integer less than or equal to x.
(i) -1 < x <0
g(x) = 1 + x – [x]
⇒ g(x) = 1 + x + 1 {∵ [x] = -1}
⇒ g(x) = 2 + x
f(g(x))= f(2 + x)
⇒ f(g(x))=1 + 2 + x-[2 + x]
⇒ f(g(x)) = 3 + x -2 – x
⇒ f(g(x)) = 1
(ii) x = 0
f(g(x)) = f(1 + x-[x])
⇒ f(g(x)) = 1 + 1 + x –[x] – [1 + x + [x]]
⇒ f(g(x)) = 2 + 0 -1
⇒ f(g(x)) = 1
(iii) x > 1
f(g(x)) = f(1 + x-[x])
⇒ f(g(x)) = f(x>0) = 1
Hence, f(g(x)) = 1 for all cases.